Möbius transformation mapping $\mathbb{D}$ into itself.

Question

I want to show that a Möbius transformation $f(z) = \frac{az + b}{cz + d}$ for $a,b,c,d \in \mathbb{C}$ maps the unit disk $\mathbb{D}$ into itself if and only if the coefficients satisfie \begin{equation*}\label{ineq} |\overline{b}d - \overline{a}c| + |ad - bc| \leq |d|^2 - |c|^2. \end{equation*}
So far I have shown that if a Möbius transformation maps $\mathbb{D}$ into itself it satisfies the inequality.
Now my idea for the implication $\Leftarrow$:
I consider the inequality above and want to show that for $|z| \leq 1 \implies$ $|f(z)| \leq 1$. Instead of showing $|f(z)| \leq 1 $ I defined for $|d| > |c|$ \begin{equation*} v = \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2}, \; R = \frac{|ad - bc|}{|d|^2 - |c|^2} \end{equation*} and want to show $|f(z) - v|^2 \leq R^2$ since by the provided inequality from $|f(z) - v|^2 \leq R^2$ follows $f(z) \in K_R(v) \subseteq \mathbb{D}$. For the case $|d| = |c|$, $f$ wouldn't be a möbius transformation.
Probelem: Now I have trouble showing the inequality above. I tried using the provided inequality show that \begin{equation*} |f(z) - v|^2 = |f(z)|^2 - 2 \Re(f(z)\overline{v}) + |v|^2 \leq |f(z)|^2 - 2 \Re(f(z)\overline{v}) + 1 - 2R + R^2 \end{equation*} but from then on I stuck by estimating \begin{align*} |f(z)|^2 &= f(z) \overline{f(z)} = \frac{|a|^2 |z|^2 + 2\Re(az \overline{b}) + |b|^2}{|c|^2 |z|^2 + 2\Re(cz \overline{d}) + |d|^2} \\ |v|^2 &= \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2}\frac{\overline{b} d - \overline{a}c}{|d|^2 - |c|^2} = \frac{|b|^2 |d|^2 - 2 \Re(b \overline{d} \overline{a}c) + |a|^2|c|^2}{(|d|^2 - |c|^2)^2} \\ &2 \Re(f(z)v) = 2 \frac{1}{|d|^2 - |c|^2}\Re \left( \frac{ab\overline{d}z - a^2 \overline{c}z - ab\overline{c} + b^2 \overline{d}}{cz + d}\right) \end{align*} or showing that \begin{align*} |f(z)|^2 - 2 \Re(f(z)\overline{v}) + 1 - 2R &\leq 0 \\ \end{align*} I made some estimates based on the fact that $|z| \leq 1$ but most of them lead nowhere. I would appreciate it if someone can give me a hint or can tell that the idea doesn't work.

Answer 1

I will be using the polarization identity $|u+v|^2=|u|^2+2\mathrm{Re}(u\bar{v})+|v|^2$ which follows from $|w|^2=w\bar{w}$ (actually, we can use $u\bar{v}$ or $\bar{u}v$ - see why?). Compare with the vector identity.

Let's figure out what $g=(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ does to the unit circle $S^1=\partial\mathbb{D}$.

Note $z\in gS^1 ~\Leftrightarrow~ g^{-1}z\in S^1 ~\Leftrightarrow~ |g^{-1}z|=1$. Since $g^{-1}$ is proportional to $(\begin{smallmatrix}d & -b \\ -c & a \end{smallmatrix})$, we get

$$ \left|\frac{\phantom{-}dz-b}{-cz+a}\right|=1 \quad\iff\quad |dz-b|^2=|cz-a|^2 $$

$$ \iff~~ |d|^2|z|^2-2\mathrm{Re}(\bar{b}dz)+|b^2| ~~=~~ |c|^2|z|^2-2\mathrm{Re}(\bar{a}cz)+|a|^2 $$

$$ \iff~~ \big(|d|^2-|c|^2\big)|z|^2+2\mathrm{Re}\big((\bar{a}c-\bar{b}d)z\big)+\big(|b|^2-|a|^2\big) ~~ =0. ​$$

Note if $|d|=|c|$ then $d=cu$ for some $u\in S^1$ so $\infty\in gS^1$ which means $gS^1$ is a line, and above the equation becomes linear which gives a line, as expected. Assume $|c|\ne|d|$ now.

To compare, observe that similarly

$$ |z-w|^2=r^2 \iff |z|^2-2\mathrm{Re}(\bar{w}z)+\big(|w|^2-r^2\big)=0. $$

Thus, normalizing the earlier quadratic and equating coefficients here yields

$$ w=\frac{\bar{b}d-\bar{a}c}{|d|^2-|c|^2}, \qquad |w|^2-r^2=\frac{|b|^2-|a|^2}{|d|^2-|c|^2}. $$

It is a fun exercise to then compute

$$ r=\sqrt{|w|^2-(|w|^2-r^2)}=\cdots=\frac{|ad-bc|}{\big||d|^2-|c|^2\big|}. $$

When is $|d|^2-|c|^2$ positive? Well, $0\le |d|^2-|c|^2$ $\,\Leftrightarrow\,$ $1\le|d/c|$ $\,\Leftrightarrow\,$ $g^{-1}\infty\not\in\mathbb{D}$ $\,\Leftrightarrow\,$ $\infty\not\in g\mathbb{D}$, or in other words when $g\mathbb{D}$ is bounded. Now we can state and prove

Theorem. $g\mathbb{D}\subseteq\mathbb{D} \iff |\bar{b}d-\bar{a}c|+|ad-bc|\le|d|^2-|c|^2$.

Note both sides of this biconditional imply $|d|-|c|^2$ is positive, so for the purposes of this proof we may assume it is. Then $g\mathbb{D}\subseteq\mathbb{D}$ is equivalent to the furthest point of $g\mathbb{D}$ (from the origin) being within $\mathbb{D}$, or in other words $|w|+r\le 1$. When we plug in the formulas for $w$ and $r$ and clear denominators, we find this is equivalent to the given inequality.

(Note that the inequality implies $0\le|d|^2-|c|^2$ which implies $g\mathbb{D}$ is bounded, so we don't have to worry about $g$ flipping $\mathbb{D}$ inside-out.)

Answer 2

I thought to share my solution if someone is interested:

\begin{align*} |f(z) - w|^2 &= \left|\frac{az + b}{cz + d} - \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2} \right|^2 \leq r^2 = \frac{|ad - bc|^2}{(|d|^2 - |c|^2)^2} \\ &\iff \left|\frac{az + b}{cz + d}(|d|^2 - |c|^2) - b \overline{d} - a\overline{c} \right|^2 \leq |ad - bc|^2 \\ &\iff \left|(az + b)(|d|^2 - |c|^2) - (cz + d)b \overline{d} - a\overline{c} \right|^2 \leq |ad - bc|^2|cz + d|^2 \\ &\iff \left| \overline{d}(azd + bd - czb -bd ) - \overline{c}(azc + bc - cza - da) \right|^2 \leq (|ad - bc| |cz + d|)^2 \\ &\iff \left| \overline{d}z(ad - cb) - \overline{c}(bc - da) \right|^2 \leq (|ad - bc| |cz + d|)^2 \\ &\iff \left| \overline{d}z + \overline{c} \right|^2 \leq |cz + d|^2 \\ &\iff \left| \overline{d \overline{z} + c } \right|^2 \leq |cz + d|^2 \\ &\iff \left| d \overline{z} + c \right|^2 \leq |cz + d|^2 \end{align*} and from $|z| \leq 1$ and $|d| \geq |c|$ we get \begin{align*} |d \overline{z} + c|^2 &= |d|^2 |z|^2 + 2 \Re(d \overline{cz}) + |c|^2 \leq |cz + d|^2 = |c|^2|z|^2 + 2 \Re(cz \overline{d}) + |d|^2 \\ &\iff |d|^2 |z|^2 + |c|^2 \leq |c|^2 |z|^2 + |d|^2 \\ &\iff |d|^2(|z|^2 - 1) \leq |c|^2(|z|^2 - 1) \\ &\iff |d|^2 \geq |c|^2 \end{align*} Therefore the inequality holds. For $d = c$ $f$ is not a mobius transformation as I already stated above in my question.